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Prelim_2_Solns

# Prelim_2_Solns - 1(20 points Given the ODE y 4y 3y = 2 sin...

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1. (20 points) Given the ODE: y + 4 y + 3 y = 2 sin x (a) Find the complementary solution. Solution: The characteristic equation is r 2 + 4 r + 3 = 0 which has roots r = - 1 and r = - 3. Therefore the complementary solution is y c = c 1 e - x + c 2 e - 3 x . (b) Find a particular solution. Solution: We note that there is no duplication between the function f ( x ) = 2 sin x and the complementary solution. We look for a solution y p of the form y p ( x ) = A cos x + B sin x. Then y p ( x ) = - A sin x + B cos x y p ( x ) = - A cos x - B sin x. Substituting this into the ODE y p + 4 y p + 3 y p = 2 sin x gives the equation - A cos x - B sin x + 4( - A sin x + B cos x ) + 3( A cos x + B sin x ) = 2 sin x or (2 A + 4 B ) cos x + (2 B - 4 A ) sin x = 2 sin x. Equating the coefficients on the right and the left gives us the following two equations for A and B . 2 A + 4 B = 0 - 4 A + 2 B = 2 which has the unique solution A = - 2 / 5 and B = 1 / 5. The particular solution is therefore y p ( x ) = - 2 5 cos x + 1 5 sin x. 1

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MATH 293, Prelim – 2 Partial Marking scheme with solution for Problem 3 Venkat Krishnan 3 a) 12 points 3 b) 8 points 3 a) kd = mg => k = mg

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